Question

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?
A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64

Answer 1

Answer: The probability that Xavier and Yvonne can solve a problem but Zelda cannot is $\frac{3}{64}$

Step-by-step explanation:

We are given:

Probability of success of Xavier, $P(S)_{Xavier}=\frac{1}{4}$

Probability of failure of Xavier, $P(F)_{Xavier}=1-\frac{1}{4}=\frac{3}{4}$

Probability of success of Yvonne, $P(S)_{Yvonne}=\frac{1}{2}$

Probability of failure of Yvonne, $P(F)_{Yvonne}=1-\frac{1}{2}=\frac{1}{2}$

Probability of success of Zelda, $P(S)_{Zelda}=\frac{5}{8}$

Probability of failure of Zelda, $P(F)_{Zelda}=1-\frac{5}{8}=\frac{3}{8}$

We need to calculate:

The probability that Xavier and Yvonne can solve the problem but Zelda cannot, we use:

$P(S)_{Xavier}\times P(S)_{Yvonne}\times P(F)_{Zelda}=\frac{1}{4}\times \frac{1}{2}\times \frac{3}{8}=\frac{3}{64}$

Hence, the probability that Xavier and Yvonne can solve a problem but Zelda cannot is $\frac{3}{64}$